Problem: $f(t) = 2t^{2}-2(h(t))$ $h(t) = 3t-1$ $g(x) = 4x^{2}+3(h(x))$ $ f(g(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-1)$ . Then we'll know what to plug into the outer function. $g(-1) = 4(-1)^{2}+3(h(-1))$ To solve for the value of $g$ , we need to solve for the value of $h(-1)$ $h(-1) = (3)(-1)-1$ $h(-1) = -4$ That means $g(-1) = 4(-1)^{2}+(3)(-4)$ $g(-1) = -8$ Now we know that $g(-1) = -8$ . Let's solve for $f(g(-1))$ , which is $f(-8)$ $f(-8) = 2(-8)^{2}-2(h(-8))$ To solve for the value of $f$ , we need to solve for the value of $h(-8)$ $h(-8) = (3)(-8)-1$ $h(-8) = -25$ That means $f(-8) = 2(-8)^{2}+(-2)(-25)$ $f(-8) = 178$